FREE BRANLIEST COMMENT AND LIKE SOMEEBODY HELP ME I BEEN WORKING ON THIS FOR TWO DAYS AND IM SLOWY DYING. A puck moves 2.35 m/s in a -22 degree direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50 degree direction. What is the magnitude of the displacement and Direction of displacement

0.805 m at 32.6°

Explanation:

Given in the x direction:

v₀ₓ = 2.35 m/s cos -22° = 2.179 m/s

vₓ = 6.42 m/s cos 50° = 4.127 m/s

t = 0.215 s

Find: Δx

Δx = ½ (v + v₀) t

Δx = ½ (4.127 m/s + 2.179 m/s) (0.215 s)

Δx = 0.678 m

Given in the y direction:

v₀ᵧ = 2.35 m/s sin -22° = -0.880 m/s

vᵧ = 6.42 m/s sin 50° = 4.918 m/s

t = 0.215 s

Find: Δy

Δy = ½ (v + v₀) t

Δy = ½ (4.918 m/s + -0.880 m/s) (0.215 s)

Δy = 0.434 m

The magnitude of the displacement is:

d² = Δx² + Δy²

d² = (0.678 m)² + (0.434 m)²

d = 0.805 m

The direction of the displacement is:

θ = tan⁻¹(Δy/Δx)

θ = tan⁻¹(0.434 m / 0.678 m)

θ = 32.6°

true is the correct answer

hope i